1. Nov 7, 2007

Dopeylass

I need help working out the RPM of the motor/gearbox and torque from the following details:
The temporary drive rigged up is running at 0.2 metres a second, which is a little faster than the speed we usually ran our broken Lenze drive. We bottle 2 sizes of glass bottle, 1 litre and 1/4 litre. The speed of our capping machine limits us to 2,000 bottles an hour (33 bottle/min). We would like to be able to go from about 0.15 metres/sec to about 4 metres/sec. I don't have any torque figures for the broken motor, but I estimate the maximum weight of bottles on the bottling line to be 30 kg and the chain conveyer is 25 kg.

If anyone knows how to work out the speed and torque from the above information please let me know as i am getting pretty desperate, as i need to order an alternative geared motor before the temporary one breaks.

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Last edited: Nov 7, 2007
2. Nov 7, 2007

FredGarvin

I think it's safe to say we need some kind of diagram showing the set up.

3. Nov 7, 2007

Dopeylass

I have photos's of the conveyor

4. Nov 14, 2007

stewartcs

If I understand your question correctly, you want to find the RPM of a motor to that will give you a tangential speed of 4 m/s right?

If so (and if my memory serves me correctly), you can find the RPM using the general formula,

RPM = (v*60)/D*Pi)

where,

v = velocity, ft/sec
D = diameter, ft

This will give you the RPM's required to obtain the desired tangential speed. The diameter is assumed to be measured from the center of the circle (or gear where the motor is coupled).

A quick Google on conveyor belt design returned lots of software that you could use instead of trying to figure it out with hand calcs.

http://www.helixtech.com.au/T5Main.htm

Hope this helps.

CS